We present here calculation of the function
\( \Phi _i ^ - \)  [Eq. (REF )]. Let us introduce the auxiliary functions:
\({\bf F} ({\bf r^{\prime }})\equiv \int _S dxdz\hspace{2.84526pt}{\bf E}_0({\bf r})\Gamma ({\bf r-r^{\prime }})\sim \exp ({-\imath \omega t})\) 
and \({\bf P} ({\bf r^{\prime }})\equiv \int _S dxdz\hspace{2.84526pt}{\bf E}_0^\ast ({\bf r})\Gamma ({\bf r-r^{\prime }})\sim \exp ({\imath \omega t})\)  (do not confuse with polarization denoted by the same letter in the main text).
Then function \( \Phi _ \alpha ({\bf r ^{\prime }})\)  Eq. (REF ) can be represented as \( \Phi _ \alpha ({\bf r ^{\prime }}) = - 2 \pi [F_ \alpha ({\bf r^{\prime }}) + P_ \alpha ({\bf r ^{\prime }})] \) 
and therefore \( \Phi ^ -_ \alpha ({\bf r ^{\prime }}) e ^ {\imath \omega t} = - 2 \pi P_ \alpha ({\bf r^{\prime }}) \) . To calculate the integral in the above expression
for \( P_ \alpha ({\bf r ^{\prime }})\) , we
note that the field \( {\bf E} _0 ^ \ast ({\bf r}) \)  satisfies the Helmholtz equation,
\(\Delta {\bf E}_0^\ast +k^2{\bf E}_0^\ast =0\) 
and, therefore,
the Kirchhoff formula [1]} can be used:
\({\bf E}_0^\ast ({\bf R})=\int _ {\partial V} \bigg [{\bf E}_0^\ast ({\bf r}){\partial \Gamma ({\bf r-R})\over \partial n}-\Gamma ({\bf r-R}){\partial {\bf E}_0^\ast ({\bf r})\over \partial n}\bigg ]d S\) 
